Simplify the following expression and state the conditions under which the simplification is valid. You can assume that $n \neq 0$. $p = \dfrac{n^2 + 11n + 18}{8n + 80} \times \dfrac{n + 10}{n^2 + 9n} $
Solution: First factor the quadratic. $p = \dfrac{(n + 9)(n + 2)}{8n + 80} \times \dfrac{n + 10}{n^2 + 9n} $ Then factor out any other terms. $p = \dfrac{(n + 9)(n + 2)}{8(n + 10)} \times \dfrac{n + 10}{n(n + 9)} $ Then multiply the two numerators and multiply the two denominators. $p = \dfrac{ (n + 9)(n + 2) \times (n + 10) } { 8(n + 10) \times n(n + 9) } $ $p = \dfrac{ (n + 9)(n + 2)(n + 10)}{ 8n(n + 10)(n + 9)} $ Notice that $(n + 10)$ and $(n + 9)$ appear in both the numerator and denominator so we can cancel them. $p = \dfrac{ \cancel{(n + 9)}(n + 2)(n + 10)}{ 8n(n + 10)\cancel{(n + 9)}} $ We are dividing by $n + 9$ , so $n + 9 \neq 0$ Therefore, $n \neq -9$ $p = \dfrac{ \cancel{(n + 9)}(n + 2)\cancel{(n + 10)}}{ 8n\cancel{(n + 10)}\cancel{(n + 9)}} $ We are dividing by $n + 10$ , so $n + 10 \neq 0$ Therefore, $n \neq -10$ $p = \dfrac{n + 2}{8n} ; \space n \neq -9 ; \space n \neq -10 $